Problem 102: Triangle Containment

Problem 102 source code

C++

#include <cstdint>
#include <fstream>
#include <iostream>
#include <sstream>
#include <stdexcept>
#include <string>
#include <vector>

namespace {

using i64 = std::int64_t;

struct Options {
    std::string file = "resources/documents/0102_triangles.txt";
    bool run_checkpoints = true;
};

struct Point {
    i64 x = 0;
    i64 y = 0;
};

bool parse_string_after_prefix(const std::string& arg,
                               const std::string& prefix,
                               std::string& value) {
    if (arg.rfind(prefix, 0U) != 0U) {
        return false;
    }
    const std::string tail = arg.substr(prefix.size());
    if (tail.empty()) {
        return false;
    }
    value = tail;
    return true;
}

bool parse_arguments(int argc, char** argv, Options& options) {
    for (int i = 1; i < argc; ++i) {
        const std::string arg(argv[i]);
        if (arg == "--skip-checkpoints") {
            options.run_checkpoints = false;
            continue;
        }
        if (parse_string_after_prefix(arg, "--file=", options.file)) {
            continue;
        }

        std::cerr << "Unknown argument: " << arg << '\n';
        return false;
    }

    return true;
}

i64 cross(const Point& a, const Point& b) {
    return a.x * b.y - a.y * b.x;
}

bool contains_origin(const Point& a, const Point& b, const Point& c) {
    const i64 c1 = cross(a, b);
    const i64 c2 = cross(b, c);
    const i64 c3 = cross(c, a);

    const bool all_positive = (c1 > 0 && c2 > 0 && c3 > 0);
    const bool all_negative = (c1 < 0 && c2 < 0 && c3 < 0);
    return all_positive || all_negative;
}

std::vector<i64> parse_csv_i64(const std::string& line) {
    std::vector<i64> values;
    std::istringstream input(line);
    std::string token;
    while (std::getline(input, token, ',')) {
        values.push_back(std::stoll(token));
    }
    return values;
}

int solve_from_text(const std::string& text) {
    std::istringstream input(text);
    std::string line;
    int count = 0;

    while (std::getline(input, line)) {
        if (line.empty()) {
            continue;
        }
        const std::vector<i64> v = parse_csv_i64(line);
        if (v.size() != 6) {
            throw std::runtime_error("Malformed triangle row");
        }

        const Point a{v[0], v[1]};
        const Point b{v[2], v[3]};
        const Point c{v[4], v[5]};
        if (contains_origin(a, b, c)) {
            ++count;
        }
    }

    return count;
}

int solve(const std::string& file_path) {
    std::ifstream input(file_path);
    if (!input) {
        throw std::runtime_error("Could not open triangles file: " + file_path);
    }

    std::ostringstream buffer;
    buffer << input.rdbuf();
    return solve_from_text(buffer.str());
}

bool run_checkpoints() {
    const std::string sample =
        "-340,495,-153,-910,835,-947\n"
        "-175,41,-421,-714,574,-645\n";

    if (solve_from_text(sample) != 1) {
        std::cerr << "Checkpoint failed for statement sample" << '\n';
        return false;
    }

    return true;
}

}  // namespace

int main(int argc, char** argv) {
    Options options;
    if (!parse_arguments(argc, argv, options)) {
        return 1;
    }
    if (options.run_checkpoints && !run_checkpoints()) {
        return 2;
    }

    try {
        std::cout << solve(options.file) << '\n';
    } catch (const std::exception& ex) {
        std::cerr << ex.what() << '\n';
        return 3;
    }

    return 0;
}

Python

# Problem 102: Triangle containment
# How many triangles in the file contain the origin?

import os

def solve():
    script_dir = os.path.dirname(os.path.abspath(__file__))
    file_path = os.path.join(script_dir, '..', 'resources', 'documents', '0102_triangles.txt')
    
    def contains_origin(x1, y1, x2, y2, x3, y3):
        # Cross product method
        d1 = x1 * (y2 - y3) - y1 * (x2 - x3) + (x2 * y3 - x3 * y2)
        d2 = 0  # sub-triangles with origin
        # Check using sign of cross products
        def sign(x):
            return (x > 0) - (x < 0)
        def cross(ax, ay, bx, by):
            return ax * by - ay * bx
        
        c1 = cross(x1, y1, x2, y2)
        c2 = cross(x2, y2, x3, y3)
        c3 = cross(x3, y3, x1, y1)
        
        has_neg = (c1 < 0) or (c2 < 0) or (c3 < 0)
        has_pos = (c1 > 0) or (c2 > 0) or (c3 > 0)
        return not (has_neg and has_pos)
    
    with open(file_path) as f:
        lines = f.readlines()
    
    count = 0
    for line in lines:
        coords = list(map(int, line.strip().split(',')))
        if contains_origin(*coords):
            count += 1
    print(count)

solve()

Java

import java.nio.file.*;
import java.util.*;

public class Euler102 {
    public static void main(String[] args) throws Exception {
        List<String> lines = Files.readAllLines(Path.of("resources/documents/0102_triangles.txt"));
        int count = 0;
        for (String line : lines) {
            String[] p = line.trim().split(",");
            if (p.length < 6)
                continue;
            int x1 = Integer.parseInt(p[0]), y1 = Integer.parseInt(p[1]), x2 = Integer.parseInt(p[2]),
                    y2 = Integer.parseInt(p[3]), x3 = Integer.parseInt(p[4]), y3 = Integer.parseInt(p[5]);
            long c1 = (long) x1 * y2 - (long) y1 * x2, c2 = (long) x2 * y3 - (long) y2 * x3,
                    c3 = (long) x3 * y1 - (long) y3 * x1;
            boolean neg = c1 < 0 || c2 < 0 || c3 < 0, pos = c1 > 0 || c2 > 0 || c3 > 0;
            if (!(neg && pos))
                count++;
        }
        System.out.println(count);
    }
}